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Solve by completing the square
(4-e)^(2)+(6-e)^(2)=20

Solve by completing the square (4-e)^(2)+(6-e)^(2)=20-example-1
User Niranga
by
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1 Answer

2 votes

Answer:


\boxed{e=8, e=2}

Explanation:

1. Expand
\large \displaystyle \left(4-e\right)^2\:+\left(6-e\right)^2

Expand each term in the expression using the identity

(a-b)^2 = a^2 -2ab+b^2

(4 - e)^2= 16-8e+e^2\\\\\left(6-e\right)^2 = 36-12e+e^2


\large \displaystyle \left(4-e\right)^2\:+\left(6-e\right)^2 = 16-8e+e^2 + 36-12e+e^2

Combining like terms we get

(16 + 36) -(8e+12e) +(e^2 + e^2) = 20


\longrightarrow 52 - 20e + 2e^2 = 20

Subtract 52 from both sides and rearrange terms:

\longrightarrow 2e^2 - 20e = -32 \\\\

Divide both sides by 2


(2e^2-20e)/(2)=(-32)/(2)


e^2-10e=-16

To rewrite in the form (x+a)² = b

we must have 2ae = -10e and therefore a = -5

So the expression

e^2-10e=-16

can be transformed by

Adding (-5)² to both sides to complete the square:


e^2-10e+\left(-5\right)^2=-16+\left(-5\right)^2


e^2-10e+\left(-5\right)^2=\left(e-5\right)^2


-16 + (-5)^2 = -16 + 25 = 9\\\\\

So we get

\left(e-5\right)^2=9


\longrightarrow e-5 = \pm √(9) \\\\e - 5 = \pm 3\\\\\\e-5 = + 3 \implies e = 5 + 3 = 8\\\\e-5 = -3 \implies e = -3 + 5 = 2\\\\

So the solution set is

\boxed{e=8, e=2}

User Raphnguyen
by
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